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Let $ (R, N) $ be a Euclidean ring. We can determine a greatest common divisor of two elements $ a, b \in R $, $ b \neq 0 $ using the Euclidean algorithm. This is done as follows:

Let $ r_{-1}=a, r_{0}=b $. By division with remainder one defines recursively $ q_{i}, r_{i} \in R $, so that $ r_{n-2}=q_{n} \cdot r_{n-1}+r_{n} $ with $ N\left(r_{n}\right)<N\left(r_{n-1}\right) $ or $ r_{n}=0 $. Now for $ m \in \mathbb{N} $ minimal with $ r_{m}=0 $, $ r_{m-1} $ is a greatest common divisor of $ a $ and $ b $.

(a) Determine in the polynomial ring $ \mathrm{Q}[x] $ a $ d \in \mathbb{Q}[x] $ with $$ (d)=\left(x^{4}-x^{2}+4 x, x^{3}-x+5\right) $$ using the Euclidean algorithm.

(b) Find polynomials $ r, s \in \mathbb{Q}[x] $ with $ d= $ $ r\left(x^{4}-x^{2}+4 x\right)+s\left(x^{3}-x+5\right) $ by back substitution in the Euclidean algorithm.

(c) Is $ \left(x^{4}-x^{2}+4 x, x^{3}-x+5\right) $ a principal ideal in $ \mathbb{Z}[x] $ ?

Attempt / Idea:

Task a) I got 5 out if you calculate it with the above algorithm.

For b), if d = 5, you get $s = -x^3+x+1$ and $r = x^2-1$.

But I am totally stuck on c), how to find out or justify this. From the idea I thought it like the known example (2,x), that is no principal ideal. So one assumes that it is a principal ideal and then looks at whether a contradiction occurs. But if I try this, it becomes quite complicated and confusing. Does anyone see how this can be justified?

clementine1001
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  • Why? In a) I calculated d = gcd$ \left(x^{4}-x^{2}+4 x, x^{3}-x+5\right) $ with the Euclidean algorithm. – clementine1001 Jan 07 '23 at 11:03
  • By ideal mod reduction (like in Euclidean algorithm) we have $,(x(fg-1),f) = (-x,f) = (x,f(0)).,$ Now use: for $,n\in \Bbb Z$ we have $(x,n)$ is principal in $\Bbb Z[x]\iff n = 0,,$ by the first linked dupe. See the 2nd linked dupe for said ideal reduction. OP is special case $,f = x^3-x+5,\ g = 1 $ – Bill Dubuque Jan 07 '23 at 14:43
  • Beware that that the argument in your accepted answer has a (fixable) flaw - see the first comment there. But better to see the linked dupe where there are more complete and general arguments (with correctness etc vetted over a decade) – Bill Dubuque Jan 07 '23 at 14:53
  • @BillDubuque there is no flaw - see my justification in comment there. – Anne Bauval Jan 07 '23 at 14:58
  • @AnneBauval No, there is a flaw - see my reply. – Bill Dubuque Jan 07 '23 at 15:01
  • ^^^ Editing slipup above: my 1st comment should say: $(x,n)$ is principal... $!!\iff! n = 0,$ or $,n\mid 1,,$ cf. 1st linked dupe. – Bill Dubuque Jan 07 '23 at 21:44

1 Answers1

2

Searching on MSE for "ideal not principal in Z[x]", I encountered none in the first 360 posts corresponding to your situation. So I found the following first proof worthwhile. (The second proof is more standard.)

In your solution for (b), you proved that $$5\in I:=\left(x^4-x^2+4 x,x^3-x+5\right)\subset\Bbb Z[x]$$ (and not only in $\Bbb Q[x]$). Now, assume by contradiction that $$I=(D(x))$$ for some $D(x)\in\Bbb Z[x].$ Then, $D(x)$ would divide $5$ in $\Bbb Z[x]$ hence it would be equal to $\pm1$ or $\pm5.$

  • It cannot be $\pm5$ because $5$ does not divide $x^3-x+5$ for instance.
  • It cannot be $\pm1$ because $\left(x^4-x^2+4 x,x^3-x+5\right)\ne\Bbb Z[x],$ since $\forall P(x)\in\left(x^4-x^2+4 x,x^3-x+5\right)\quad5\mid P(0).$

Another proof, less in the spirit of your exercise, is to track the Euclidean algorithm to find simpler generators for your idéal (it partially corresponds to Bill's second duplicate): $$\begin{align}\left(x^4-x^2+4 x,x^3-x+5\right)&=\left(x^4-x^2+4 x-x(x^3-x+5),x^3-x+5\right)\\&=\left(-x,x^3-x+5\right)\\ &=\left(-x,x^3-x+5-x(x^2-1)\right)\\ &=(x,5)\end{align}$$ and conclude like in your "known example (2,x)" (= Bill's first "duplicate").

Anne Bauval
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  • This is not correct as written since the OP proved that the ideal $= (5)$ in $\Bbb Q[x],$ not in $\Bbb Z[x].\ $ – Bill Dubuque Jan 07 '23 at 14:52
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jan 07 '23 at 14:52
  • No, because the exercise steps $(a)$ & $(b)$ work in $\Bbb Q[x]$ so one needs to explicitly justify any stronger claims (in $\Bbb Z[x]),$ for the proot to be correct. – Bill Dubuque Jan 07 '23 at 14:59
  • OK It was correct but I added a small detail which was obvious to me, and seemingly to the OP. – Anne Bauval Jan 07 '23 at 15:03
  • You're still missing a crucial point, Your argument depends crucially on the fact that the OP's $,r,s\ \in \Bbb Z[x],,$ whereas the exercise only asked to find $,r,s\in \Bbb Q[x].\ \ $ – Bill Dubuque Jan 07 '23 at 15:11
  • We obviously disagree on pedagogical matters here so It's not worth going any further (and by now the point should be clear to the OP, from the above discussion and your edit). – Bill Dubuque Jan 07 '23 at 15:31
  • I am often ready to delete my answers when duplicates are found, but not this time. Here is why: The first part of my answer is more in the spirit of the OP's exercise than the alleged duplicates. The second part is easier to read than the 2nd duplicate (and more adapted since that duplicate focusses on ideals of the form $(x-a,\dots)$ a priori. The 1st duplicate is useless: the OP already knows it. As for the OP, I think everything was already clear to her. – Anne Bauval Jan 07 '23 at 15:33
  • Again we'll have to (strongly) disagree. Note also the linked dupes contain more general arguments than anything above. That's the point of abstract dupes - to teach how to fish vs. serving up fish on silver platters ad infinitum. Everything in your answer has already been posted here many tens (if not hundreds) of times already. – Bill Dubuque Jan 07 '23 at 15:40